To solve the given problem:

Given Ellipse Equation:

$\frac{x^2}{9} + \frac{y^2}{4} = 1$

This is in the standard form of the ellipse equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where:

• $a^2 = 9$, so $a = 3$
• $b^2 = 4$, so $b = 2$

Since $a^2 > b^2$, this is a horizontal ellipse with the following characteristics:

• Center: $(0, 0)$ (no shifts in the equation).
• Vertices: These are located at $(\pm a, 0)$, which are $(\pm 3, 0)$, or: $(-3, 0) \text{ and } (3, 0)$
• Foci: To find the foci, use the formula $c = \sqrt{a^2 - b^2}$: $c = \sqrt{9 - 4} = \sqrt{5}$ The foci are at $(\pm c, 0)$, which are: $(-\sqrt{5}, 0) \text{ and } (\sqrt{5}, 0)$

Final Answer:

• Vertices: $(-3, 0), (3, 0)$
• Foci: $(- \sqrt{5}, 0), (\sqrt{5}, 0)$

Would you like a graph or further details on this problem? Let me know if you have questions.

Here are 5 related questions to expand your understanding:

1. How would the equation change if the ellipse were vertical instead of horizontal?
2. What are the properties of a hyperbola and how do they differ from an ellipse?
3. How is the eccentricity of an ellipse calculated, and what does it represent?
4. If the center of the ellipse were shifted to $(h, k)$, how would the equation change?
5. How can you determine the length of the major and minor axes from the given equation?

Tip: Always start by identifying whether $a^2$ or $b^2$ is larger to determine if the ellipse is horizontal or vertical!